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**LIAL HORNSBY SCHNEIDER**

COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

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**Inequalities 1.7 Linear Inequalities Three-Part Inequalities**

Quadratic Inequalities Rational Inequalities

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**Properties of Inequality**

For real numbers a, b and c: If a < b, then a + c < b + c, If a < b and if c > 0, then ac < bc, If a < b and if c < 0, then ac > bc. Replacing < with >, ≤ , or ≥ results in similar properties.

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Motion Problems Note Multiplication may be replaced by division in properties 2 and 3. Always remember to reverse the direction of the inequality symbol when multiplying or dividing by a negative number.

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**Linear Inequality in One Variable**

A linear inequality in one variable is an inequality that can be written in the form where a and b are real numbers with a ≠ 0. (Any of the symbols ≥, <, or ≤ may also be used.)

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**Don’t forget to reverse the symbol here.**

SOLVING A LINEAR INEQUALITY Example 1 Solve Solution Subtract 5. Divide by – 3; reverse direction of the inequality symbol when multiplying or dividing by a negative number. Don’t forget to reverse the symbol here.

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**Set Interval Notation Graph**

Type of Interval Set Interval Notation Graph Open interval {x x > a} {x a < x < b} {x x < b} (a, ) (a, b) (– , b) Other interval {x x ≥ a} {x a < x ≤ b} {x a ≤ x < b} {x x≤ b} [a, ) (a, b] [a, b) (– , b] ( a ( a b ( b [ a ( ] a b [ ) a b ] b

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**Set Interval Notation Graph**

Type of Interval Set Interval Notation Graph Closed interval {xa ≤ x ≤ b} [a, b] Disjoint interval {xx < a or x > b} (– , a) (b, ) All real numbers {xx is a real numbers} (– , ) [ ] a b ( a b

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**SOLVING A LINEAR INEQUALITY**

Example 2 Solve 4 – 3x ≤ 7 + 2x. Give the solution set in interval notation and graph it. Solution Subtract 4. Subtract 2x.

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**SOLVING A LINEAR INEQUALITY**

Example 2 Solve 4 – 3x ≤ 7 + 2x. Give the solution set in interval notation and graph it. Solution Divide by –5; reverse the direction of the inequality symbol. [

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**Solve – 2 < 5 + 3x < 20. Solution**

SOLVING A THREE-PART INEQUALITY Example 3 Solve – 2 < 5 + 3x < 20. Solution Subtract 5 from each part. Divide each part by 3.

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**If the revenue and cost of a certain product are given by**

FINDING THE BREAK-EVEN POINT Example 4 If the revenue and cost of a certain product are given by where x is the number of units produced and sold, at what production level does R at least equal C? Solution Set R ≥ C and solve for x.

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**Solution Set R ≥ C and solve for x.**

FINDING THE BREAK-EVEN POINT Example 4 Solution Set R ≥ C and solve for x. Substitute. Subtract 2x. Divide by 2. The break-even point is at x = This product will at least break even only if the number of units produced and sold is in the interval [500, ).

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**Quadratic Inequalities**

A quadratic inequality is an inequality that can be written in the form for real numbers a, b, and c with a ≠ 0. (The symbol < can be replaced with >, ≤, or ≥.)

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**Solving a Quadratic Inequality**

Step 1 Solve the corresponding quadratic equation. Step 2 Identify the intervals determined by the solutions of the equation. Step 3 Use a test value from each interval to determine which intervals form the solution set.

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**Step 1 Find the values of x that satisfy x2 – x – 12 = 0**

SOLVING A QUADRATIC INEQULITY Example 5 Solve Solution Step 1 Find the values of x that satisfy x2 – x – 12 = 0 Corresponding quadratic equation Factor. or Zero-factor property

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**Solve Solution Step 1 or or SOLVING A QUADRATIC INEQULITY Example 5**

Zero-factor property or

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**SOLVING A QUADRATIC INEQULITY**

Example 5 Step 2 The two numbers – 3 and 4 divide the number line into three intervals. If a value in Interval A makes the polynomial negative, then all values in Interval A will make the polynomial negative. – 3 4 Interval A Interval B Interval C (–, – 3) (– 3, 4) (4, )

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**Step3 Choose a test value from each interval.**

SOLVING A QUADRATIC INEQULITY Example 5 Step3 Choose a test value from each interval. Interval Test Value Is x2 – x – 12< 0 True or False? A: (– , – 3) – 4 (– 4)2 – 4(– 4) – 12 < ? 8 < False B: (– 3, 4) 02 – 0 – 12 < ? – 12 < True C: (4, ) 5 52 – 5 – 12 < ? Since the values in Interval B make the inequality true, the solution set is (– 3, 4).

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**Step 1 Find the values of x that satisfy this equation.**

SOLVING QUADRATIC INEQULAITY Example 6 Solve Solution Step 1 Find the values of x that satisfy this equation. Corresponding quadratic equation Factor. or Zero-factor property

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**Step 1 Find the values of x that satisfy this equation.**

SOLVING QUADRATIC INEQULAITY Example 6 Solve Solution Step 1 Find the values of x that satisfy this equation. or Zero-factor property or

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**Step 2 The values form the intervals on the number line.**

SOLVING QUADRATIC INEQULAITY Example 6 Solve Solution Step 2 The values form the intervals on the number line. – 4 3/2 Interval A Interval B Interval C (–, – 4) (– 4, 3/2) (3/2, )

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**Step3 Choose a test value from each interval.**

SOLVING QUADRATIC INEQULAITY Example 6 Step3 Choose a test value from each interval. Interval Test Value Is 2x2 + 5x – 12 ≥ 0 True or False? A: (– , – 4) – 5 2(– 4)2 +5(– 5) – 12 ≥ 0 ? 13 ≥ 0 True B: (4, 3/2) 2(0)2 +5(0) – 12 ≥ 0 ? – 12 ≥ 0 False C: (3/2, ) 2 2(2)2 + 5(2) – 12 ≥ 0 ? 6 ≥ True The values in Intervals A and B make the inequality true, the solution set is the union of the intervals.

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Inequalities Note Inequalities that use the symbols < and > are called strict inequalities; ≤ and ≥ are used in nonstrict inequalities. The solutions of the equation in Example 5 were not included in the solution set since the inequality was a strict inequality. In Example 6, the solutions of the equation were included in the solution set because of the nonstrict inequality.

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**When will the projectile be greater than 80 ft above ground level?**

SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 7 If a projectile is launched from ground level with an initial velocity of 96 ft per sec, its height in feet t seconds after launching is s feet, where When will the projectile be greater than 80 ft above ground level?

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**Reverse direction of the inequality symbol.**

SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE Example 7 Solution Set s greater than 80. Subtract 80. Divide by – 16. Reverse direction of the inequality symbol. Now solve the corresponding equation.

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**Solution or SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE**

Example 7 Solution Factor. or Zero-factor property

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**SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE**

Example 7 Solution Interval C 1 5 Interval A Interval B (– , 1) (1, 5) (5, ) Use the procedure of Examples 5 and 6 to determine that values in Interval B, (1, 5) , satisfy the inequality. The projectile is greater than 80 ft above ground level between 1 and 5 sec after it is launched.

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**Solving a Rational Inequality**

Step 1 Rewrite the inequality, if necessary, so that 0 is on one side and there is a single fraction on the other side. Step 2 Determine the values that will cause either the numerator or the denominator of the rational expression to equal 0. These values determine the intervals of the number line to consider.

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**Solving a Rational Inequality**

Step 3 Use a test value from each interval to determine which intervals form the solution set. A value causing the denominator to equal zero will never be included in the solution set. If the inequality is strict, any value causing the numerator to equal zero will be excluded; if nonstrict, any such value will be included.

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**by multiplying both sides by x + 4 to obtain **

Caution Solving a rational inequality such as by multiplying both sides by x + 4 to obtain 5 ≥ x + 4 requires considering two cases, since the sign of x + 4 depends on the value of x. If x + 4 were negative, then the inequality symbol must be reversed. The procedure described in the next two examples eliminates the need for considering separate cases.

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**Solve Solution Step 1 Example 8 SOLVING A RATIONAL INEQUALITY**

Subtract 1 so that 0 is on one side. Use x + 4 as the common denominator.

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**Note the careful use of parentheses.**

SOLVING A RATIONAL INEQUALITY Example 8 Solve Solution Step 1 Use x + 4 as the common denominator. Note the careful use of parentheses. Write as a single fraction. Combine terms in the numerator; be careful with signs.

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**SOLVING A RATIONAL INEQUALITY**

Example 8 Solve Solution Step 2 The quotient possibly changes sign only where x-values make the numerator or denominator 0. This occurs at or or

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**Solve Solution Step 2 Example 8 SOLVING A RATIONAL INEQUALITY**

– 4 1 Interval A Interval B Interval C (–, – 4) (– 4, 1) (1, )

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**Step3 Choose test values.**

SOLVING A RATIONAL INEQUALITY Example 8 Step3 Choose test values. Interval Test Value A: (– , – 4) – 5 B: (– 4, 1) C: (1, ) 2 The values in the interval (– 4, 1) satisfy the original inequality. The value 1 makes the nonstrict inequality true, so it must be included in the solution set. Since – 4 makes the denominator 0, it must be excluded. The solutions set is (– 4, 2].

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Caution As suggested by Example 8, be careful with the endpoints of the intervals when solving rational inequalities.

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**Solve Solution SOLVING A RATIONAL INEQULAITY Example 9 Subtract 5.**

Common denominator is 3x + 4. Write as a single fraction.

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**Solve Solution SOLVING A RATIONAL INEQULAITY Example 9**

Write as a single fraction. Distributive property Be careful with signs. Combine terms in the numerator.

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**SOLVING A RATIONAL INEQULAITY**

Example 9 Solve Solution Set the numerator and denominator equal to 0 and solve the resulting equations to get the values of x where sign changes may occur. or or

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**Solve Solution SOLVING A RATIONAL INEQULAITY Example 9 Interval A**

Interval B Interval C

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**Solution SOLVING A RATIONAL INEQULAITY Example 9**

Now choose from the intervals and verify that: – 2 from Interval A makes the inequality true; – 1.5 from Interval B makes the inequality false; from Interval C makes the inequality true. Because of the < symbol, neither endpoint satisfies the inequality, so the solution set is

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